Circle

Exercise 11.3

Question 7: Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

semi circle

Solution: Given, diameter of the semicircle = 10cm.

Circumference of the circle `= π D`
`= 3.14 xx 10 = 31.4`

So, perimeter of semicircle `= 31.4 ÷ 2 = 25.7` cm

Perimeter of the given figure `= 25.7 + 10 = 35.7` cm


Question 8: Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs 15/m^2. (Take π = 3.14)

Solution: To calculate the cost of polishing of circular table, we have to calculate its area first of all.Here, Given diameter of the circular table-top = 1.6 m. Hence, r = 0.8 m

Area `= π r^2`
`= 3.14 xx 0.8 xx 0.8 = 2.0096` sq m

Cost of polishing = Rate `xx` Area
`= Rs. 15 xx 2.0096 = Rs. 30.144`


Question 9: Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square?

Solution: The length of the wire = 44 cm = circumference of the circle, Radius = ?
When the same wire is bent into the shape of square,
Perimeter of the square = length of the wire = 44 cm
To know that which shape will encloses more area, we have to calculate their areas.

Circumference `= 2 π r`
Or, `44 = 2 xx (22)/(7) xx\ r`
Or, `r = 44 xx (7)/(44) = 7` cm

Now, Area `= π r^2`
`= (22)/(7) xx 7 xx 7 = 22 xx 7 = 154` sq cm

When the same wire is bent to the shape of a square

perimeter of a square `= 4 xx \text(side)`
Or, `44 = 4 xx \text(side)`
Or, side `= 44 ÷ 4 = 11` cm

Area of a square = side2`= 11^2= 121` cm2

It is clear; area of the circle > Area of square


Question 10: From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet.

semi circle

Solution: Radius of circular card sheet = 14 cm
Radius of each circle which has been removed = 3.5 cm
Length of the rectangle which has been removed = 3cm
Breadth of the rectangle which has been removed = 1 cm

Area of circle `= π r^2`
`= (22)/(7) xx 14 xx 14`
`= 22 xx 2 xx 14 = 616` sq cm

Now,Area of the circle which is removed `= π r^2`
`= (22)/(7) xx 7/2 xx 7/2`
`= 11 xx 7/2 = 38.5` sq cm
Area of two such circles `= 2 xx 38.5= 77` cm2

Area of the rectangle which has been removed = length `xx` breadth
`= 3xx1= 3` cm2

Total area of the shapes which has been removed = Area of two circles + area of rectangle
= 77cm2+ 3 cm2= 80 cm2

Area left after removing the two circles and one rectangle
= Area of circular sheet — Total area of the shapes which has been removed
= 616 cm2 — 80 cm2 = 536 sq cm



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